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CHAPTER 5
IMPERFECTIONS IN SOLIDS
PROBLEM SOLUTIONS
Vacancies and Self-Interstitials
5.1 Calculate the fraction of atom sites that are vacant for copper at its melting temperature of 1084°C
(1357 K). Assume an energy for vacancy formation of 0.90 eV/atom.
Solution
In order to compute the fraction of atom sites that are vacant in copper at 1357 K, we must employ Equation
5.1. As stated in the problem, Qv = 0.90 eV/atom. Thus,
Nv
N= exp
Qv
kT
= exp
0.90 eV /atom
(8.62 105 eV /atom - K) (1357 K)
= 4.56 10-4
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5.2 Calculate the energy for vacancy formation in silver, given that the equilibrium number of vacancies
at 800°C (1073 K) is 3.6 × 1023
m–3
. The atomic weight and density (at 800°C) for silver are, respectively, 107.9
g/mol and 9.5 g/cm3.
Solution
This problem calls for the computation of the activation energy for vacancy formation in silver. Upon
examination of Equation 5.1, all parameters besides Qv are given except N, the total number of atomic sites.
However, N is related to the density, (), Avogadro's number (NA), and the atomic weight (A) according to Equation
5.2 as
N = N APb
APb
= (6.02 1023 atoms /mol)(9.5 g /cm3)
107.9 g /mol
= 5.30 1022 atoms/cm3 = 5.30 1028 atoms/m3
Now, taking natural logarithms of both sides of Equation 5.1,
ln Nv = ln N Qv
kT
and, after some algebraic manipulation
Qv = kT ln N v
N
Now, inserting values for the parameters given in the problem statement leads to
Qv = (8.62 10-5 eV/atom- K) (800C + 273 K) ln 3.60 1023 m3
5.30 1028 m3
= 1.10 eV/atom
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Impurities in Solids
5.5 Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated,
for several elements; for those that are nonmetals, only atomic radii are indicated.
Atomic Crystal Electro-
Element Radius (nm) Structure negativity Valence
Ni 0.1246 FCC 1.8 +2
C 0.071
H 0.046
O 0.060
Ag 0.1445 FCC 1.9 +1
Al 0.1431 FCC 1.5 +3
Co 0.1253 HCP 1.8 +2
Cr 0.1249 BCC 1.6 +3
Fe 0.1241 BCC 1.8 +2
Pt 0.1387 FCC 2.2 +2
Zn 0.1332 HCP 1.6 +2
Which of these elements would you expect to form the following with nickel:
(a) A substitutional solid solution having complete solubility
(b) A substitutional solid solution of incomplete solubility
(c) An interstitial solid solution
Solution
In this problem we are asked to cite which of the elements listed form with Ni the three possible solid
solution types. For complete substitutional solubility the following criteria must be met: 1) the difference in atomic
radii between Ni and the other element (R%) must be less than ±15%, 2) the crystal structures must be the same, 3)
the electronegativities must be similar, and 4) the valences should be the same, or nearly the same. Below are
tabulated, for the various elements, these criteria.
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Crystal Electro-
Element R% Structure negativity Valence
Ni FCC 2+
C –43
H –63
O –52
Ag +16 FCC +0.1 1+
Al +15 FCC -0.3 3+
Co +0.6 HCP 0 2+
Cr +0.2 BCC -0.2 3+
Fe -0.4 BCC 0 2+
Pt +11 FCC +0.4 2+
Zn +7 HCP -0.2 2+
(a) Pt is the only element that meets all of the criteria and thus forms a substitutional solid solution having
complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal
structure, and thus display complete solid solubility at these temperatures.
(b) Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals
have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Ni are
greater than ±15%, and/or have a valence different than 2+.
(c) C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly
smaller than the atomic radius of Ni.
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5.6 (a) Suppose that CaO is added as an impurity to Li2O. If the Ca2+
substitutes for Li+, what kind of
vacancies would you expect to form? How many of these vacancies are created for every Ca2+
added?
(b) Suppose that CaO is added as an impurity to CaCl2. If the O2–
substitutes for Cl–, what kind of
vacancies would you expect to form? How many of these vacancies are created for every O2–
added?
Solution
(a) For Ca2+ substituting for Li+ in Li2O, lithium vacancies would be created. For each Ca2+ substituting
for Li+, one positive charge is added; in order to maintain charge neutrality, a single positive charge may be
removed. Positive charges are eliminated by creating lithium vacancies, and for every Ca2+ ion added, a single
lithium vacancy is formed.
(b) For O2- substituting for Cl- in CaCl2, chlorine vacancies would be created. For each O2- substituting
for a Cl-, one negative charge is added; negative charges are eliminated by creating chlorine vacancies. In order to
maintain charge neutrality, one O2- ion will lead to the formation of one chlorine vacancy.
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Specification of Composition
5.7 What is the composition, in atom percent, of an alloy that consists of 92.5 wt% Ag and 7.5 wt% Cu?
Solution
In order to compute composition, in atom percent, of a 92.5 wt% Ag-7.5 wt% Cu alloy, we employ
Equation 5.9 as
CAg' =
CAgACu
CAgACu CCu AAg
100
= (92.5)(63.55 g /mol)
(92.5)(63.55 g /mol) (7.5)(107.87g /mol) 100
= 87.9 at%
CCu' =
CCu AAg
CAgACu CCu AAg
100
= (7.5)(107.87 g /mol)
(92.5)(63.55 g /mol) (7.5)(107.87g /mol) 100
= 12.1 at%
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5.8 Calculate the composition, in weight percent, of an alloy that contains 105 kg of iron, 0.2 kg of
carbon, and 1.0 kg of chromium.
Solution
The concentration, in weight percent, of an element in an alloy may be computed using a modified form of
Equation 5.6. For this alloy, the concentration of iron (CFe) is just
CFe = mFe
mFe mC mCr
100
= 105 kg
105 kg 0.2 kg 1.0 kg 100 = 98.87 wt%
Similarly, for carbon
CC = 0.2 kg
105 kg 0.2 kg 1.0 kg 100 = 0.19 wt%
And for chromium
CCr = 1.0 kg
105 kg 0.2 kg 1.0 kg 100 = 0.94 wt%
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5.16 Iron and vanadium both have the BCC crystal structure, and V forms a substitutional solid solution for
concentrations up to approximately 20 wt% V at room temperature. Compute the unit cell edge length for a 90 wt%
Fe–10 wt% V alloy.
Solution
First of all, the atomic radii for Fe and V (using the table inside the front cover) are 0.124 and 0.132 nm,
respectively. Also, using Equation 3.5 it is possible to compute the unit cell volume, and inasmuch as the unit cell is
cubic, the unit cell edge length is just the cube root of the volume. However, it is first necessary to calculate the
density and average atomic weight of this alloy using Equations 5.13a and 5.14a. Inasmuch as the densities of iron
and vanadium are 7.87g/cm3 and 6.10 g/cm3, respectively, (as taken from inside the front cover), the average density
is just
ave = 100
CV
V
CFe
Fe
= 100
10 wt%
6.10 g /cm3
90 wt%
7.87 g /cm3
= 7.65 g/cm3
And for the average atomic weight
Aave = 100
CV
AV
CFe
AFe
= 100
10 wt%
50.94 g /mole
90 wt%
55.85 g /mol
= 55.32 g/mol
Now, VC is determined from Equation 3.5 as
VC = nAave
aveNA
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= (2 atoms /unit cell)(55.32 g /mol)
(7.65 g /cm3)(6.02 1023 atoms /mol)
= 2.40 10-23 cm3/unit cell
And, finally
a = (VC )1/3
= (2.40 1023cm3/unit cell)1/3
= 2.89 x 10-8 cm = 0.289 nm
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Interfacial Defects
5.17 For an FCC single crystal, would you expect the surface energy for a (100) plane to be greater or
less than that for a (111) plane? Why? (Note: You may want to consult the solution to Problem W3.46 at the end
of Chapter 3.)
Solution
The surface energy for a crystallographic plane will depend on its packing density [i.e., the planar density
(Section 3.11)]—that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the
more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the
solution to Problem W3.46, planar densities for FCC (100) and (111) planes are
1
4R2 and
1
2R2 3, respectively—
that is
0.25
R2 and
0.29
R2 (where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have
the lower surface energy.
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5.18 (a) For a given material, would you expect the surface energy to be greater than, the same as, or less
than the grain boundary energy? Why?
(b) The grain boundary energy of a small-angle grain boundary is less than for a high-angle one. Why is
this so?
Solution
(a) The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms
on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore,
there will be fewer unsatisfied bonds along a grain boundary.
(b) The small-angle grain boundary energy is lower than for a high-angle one because more atoms bond
across the boundary for the small-angle, and, thus, there are fewer unsatisfied bonds.
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Grain Size Determination
5.20 (a) Employing the intercept technique, determine the average grain size for the steel specimen whose
microstructure is shown in Figure 10.29(a); use at least seven straight-line segments.
(b) Estimate the ASTM grain size number for this material.
Solution
(a) This portion of the problem calls for a determination of the average grain size of the specimen which
microstructure is shown in Figure 10.29(a). Seven line segments were drawn across the micrograph, each of which
was 60 mm long. The average number of grain boundary intersections for these lines was 6.3. Therefore, the
average line length intersected is just
60 mm
6.3 = 9.5 mm
Hence, the average grain diameter, d, is
d = ave. line length intersected
magnification =
9.5 mm
90= 0.106 mm
(b) This portion of the problem calls for us to estimate the ASTM grain size number for this same material.
The average grain size number, n, is related to the number of grains per square inch, N, at a magnification of 100
according to Equation 5.19. However, the magnification of this micrograph is not 100x, but rather 90.
Consequently, it is necessary to use Equation 5.20
N MM
100
2
2n1
where NM = the number of grains per square inch at magnification M, and n is the ASTM grain size number. Taking
logarithms of both sides of this equation leads to the following:
log N M 2 log M
100
(n 1) log 2
Solving this expression for n gives
n
log N M 2 logM
100
log 2 1
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From Figure 10.29(a), NM is measured to be approximately 4, which leads to
n
log 4 2 log90
100
log 2 1
= 2.7
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DESIGN PROBLEMS
5.D2 Gallium arsenide (GaAs) and indium arsenide (InAs) both have the zinc blende crystal structure and
are soluble in each other at all concentrations. Determine the concentration in weight percent of InAs that must be
added to GaAs to yield a unit cell edge length of 0.5820 nm. The densities of GaAs and InAs are 5.316 and 5.668
g/cm3, respectively.
Solution
This problem asks that we determine the concentration (in weight percent) of InAs that must be added to
GaAs to yield a unit cell edge length of 0.5820 nm. The densities of GaAs and InAs were given in the problem
statement as 5.316 and 5.668 g/cm3, respectively. To begin, it is necessary to employ Equation 3.6, and solve for
the unit cell volume, VC, for the InAs-GaAs alloy as
VC = n' Aave
aveNA
where Aave and ave are the atomic weight and density, respectively, of the InAs-GaAs alloy. Inasmuch as both of
these materials have the zinc blende crystal structure, which has cubic symmetry, VC is just the cube of the unit cell
length, a. That is
VC = a3 = (0.5820 nm)3
= (5.820 108 cm)3 1.971 1022 cm3
It is now necessary to construct expressions for Aave and ave in terms of the concentration of indium arsenide,
CInAs using Equations 5.14a and 5.13a. For Aave we have
Aave = 100
CInAs
AInAs
(100 CInAs)
AGaAs
= 100
CInAs
189.74 g /mol
(100 CInAs)
144.64 g /mol
whereas for ave
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ave = 100
CInAs
InAs
(100 CInAs)
GaAs
= 100
CInAs
5.668 g /cm3
(100 CInAs)
5.316 g /cm3
Within the zinc blende unit cell there are four formula units, and thus, the value of n' in Equation 3.6 is 4; hence, this
expression may be written in terms of the concentration of InAs in weight percent as follows:
VC = 1.971 x 10-22 cm3
= n' Aave
aveN A
=
(4 fu /unit cell)100
CInAs
189.74 g /mol
(100 CInAs)
144.64 g /mol
100
CInAs
5.668 g /cm3
(100 CInAs)
5.316 g /cm3
(6.02 1023 fu /mol)
And solving this expression for CInAs leads to CInAs = 46.1 wt%.